\documentclass[12pt]{article}
\usepackage[margin=0.5in]{geometry}

%http://tex.stackexchange.com/questions/43743/how-to-reduce-line-space-leading-within-an-enumerate-environment
%\usepackage{paralist}
\usepackage{lipsum}
\usepackage{enumitem}
\usepackage{bm}
%\setlist[enumerate]{itemsep=27mm}
%\setlist{noitemsep}
\setlist{itemsep=0.5mm}
\setenumerate[0]{label=(\alph*)}


\begin{document}
%\pagenumbering{gobble}

\large{Name:} \hfill    \large{Probability for Scientists, Fall 2013}

\large{Collaborator(s):} \hfill    \large{ Bio 409 / Bio 509 / Stat 479 } 

\hfill    \large{ Lab 1 (35pts), Due 3 Sep 2013 }

\large{You don't need a calculator for this assignment.  Please show your work,
and leave answers as fractions.}

\section{(5) Ask a question:}
Email a question to the class address (probforsci@x14n.org).  The question can
be about this lab, or anything from class or the readings.  The subject of the
email should summarize the question.


\section{(5) Solve for the following:}
    \begin{enumerate}
        \item $\frac{3^4}{2^5}$ \hfill \boldmath{$81/32$}
        \item $(\frac{2}{3})^4$ \hfill \boldmath{$16/81$}
        \item $5!$ \hfill \boldmath{$120$}
        \item $\frac{1,000!}{998!}$ \hfill \boldmath{$1,000 * 999 = 999,000$}
        \item $\frac{(10^5)!}{(10^5-1)!}$ \hfill \boldmath{$10^5$}
    \end{enumerate}


\section{(5) Write the following as fractions:}
    \begin{enumerate}
        \item Nine times out of ten. \hfill \boldmath{$9/10$}
        \item Never.  \hfill \boldmath{$0$}
        \item Always.  \hfill \boldmath{$1$}
        \item Even odds.  \hfill \boldmath{$1/2$}
        \item Once in a blue moon.  
    \end{enumerate}

\textbf{Assuming that a blue moon lasts one
day, \boldmath{$1/991$}.  Alternately, \boldmath{$1/N$} for N large (this second
answer is closer to what we mean when we use the phrase in speech).}


\section{(20) Mating pairs:}
    \begin{enumerate}
        \item A nuclear apocalypse has ravaged the globe leaving 10 surviving pigeons, only 4
of which are capable of reproduction.  What is the probability that at least one
breeding pair of pigeons has survived to repopulate the world?
        ~\\
\textbf{Only 4 pigeons are capable of reproduction, so we only care about them.
Each bird can be male or female, so we have \boldmath{$2^4 = 16$} possible
combinations.  2 of those combinations lead to {\em no} reproduction (all M or
all F).  Thus, \boldmath{$14/16 = 7/8$} combinations have at least one breeding
pair.}
    
        \item What assumptions did you make to arrive at this probability?
        ~\\
\textbf{Both sexes are equally common, and equally likely to have survived the
apocalypse capable of reproduction.  That post-apocalyptic pigeons aren't too
picky in mate choice.}

        \item In the above scenario, assume that pigeons have 5 possible sexes
(some fungi, for example, have more than 2 sexes).  Assume that each sex 
was equally abundant before the apocalypse, and equally likely to survive.  Also
assume that members of each sex can mate with any member of a sex other than its
own.  Now what is the probability of at least one breeding pair?
        ~\\
\textbf{ We still have 4 birds, but now we have \boldmath{$5^4 = 625$} possible
combinations.  5 of those combinations lead to {\em no} reproduction (all the 
same sex).  Thus, \boldmath{$620/625 = 124/125$} combinations have at least one breeding
pair.}
        ~\\

        \item Does this suggest a benefit from having more than 2 sexes?  Can
you think of a scenario where this would be helpful?

\textbf{Consider a species where encounters between individuals of that species are
rare.  In the case of only 2 sexes, 1/2 of encounters between 2
individuals will be sexually incompatibility.  As the number of sexes increases,
the probability that 2 (or few) individuals are of the same sex rapidly
decreases, increasing the number of mating opportunities.
}
    \end{enumerate}

\end{document}
